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3 years ago

I need help with 2 questions . Please help c:

1 answer:

8 0

A) According to the nebular theory, the Solar System formed from a huge gaseous nebula which at a certain point was perturbated. Atoms and molecules started colliding, forming planetesimals (a sort of big rocks). The planetesimals were attracted to each other by gravity, forming bigger warm almost spherical objects called protoplanets, which at the end cooled down forming planets.
Therefore the correct answer is "all of the above".

b) The planets closer to the Sun were (and still are) subject to higher temperatures, due to their close distance to the Sun. In these conditions, rocky materials undergo condensation, while iced gaseous materials undergo vaporization. In the outer parts of the Solar System temperatures are too low to allow these transformations.
The correct answer is again "all of the above".

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In a pig caller can produce a sound intensity level of 107 dB. How many pig callers would be needed to generate an intensity lev

myrzilka [38]

Answer:

20 pig callers

Explanation:

Given that:

A pig caller produced intensity level of a sound = 107 dB

To find how many pig callers required to generate an intensity level of 120 dB;

we have:

120 dB - 107 dB = 13 dB

Taking the logarithm function;

$10 \ log \bigg(\dfrac{I}{I_o} \bigg) = 13 \ dB$

where;

$I_o$ = initial intensity

$log \bigg(\dfrac{I}{I_o} \bigg) = 1.3$

$\dfrac{I}{I_o}= 10^{1.3 }$

I = 19.95 $I_o$

I ≅ 20 pig callers

6 0

3 years ago

Charge A and charge B are 3.00 m apart, and charge A is +1.44 C and charge B is +3.10 C. Charge C is located between them at a c

Sidana [21]

Answer:

the distance from charge A to C is r₁₃= 1.216 m

Explanation:

following Coulomb's law , the force exerted by 2 point charges between themselves is:

F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant

since C ( denoted as 3) is at equilibrium

F₁₃=F₂₃

k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²

q₁/r₁₃²=q₂/r₂₃²

r₁₃²/q₁=r₂₃²/q₂

r₂₃=r₁₃*√(q₂/q₁)

since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have

r₁₃+r₂₃=d=r₁₂

r₁₃+r₁₃*√(q₂/q₁)=d

r₁₃*(1+√(q₂/q₁))=d

r₁₃=d/(1+√(q₂/q₁))

replacing values

r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m

thus the distance from charge A to C is r₁₃= 1.216 m

7 0

3 years ago

What accounts for most of the mass in the universe?

aev [14]

Around 80 percent of the mass of the universe is made up material known as "Dark matter". It does not emit light or energy but the influence of it can be detected or observed gravitationally. Motions of stars and galaxy tell us how much mater there is, but somehow the speed of rotation of galaxy does not add up to its mass alone, there is a certain amount of matter really not accounted for. Dark matter maybe made up of non-baryonic matter, or perhaps what scientist called the WIMPS or (weakly interacting massive particles.)

5 0

3 years ago

Twin space probes have a mass of 722 kg each. If the gravitational force between the two space probes is 8.61

IceJOKER [234]

Answer:200×10^5 meters

Explanation:

6 0

3 years ago

Read 2 more answers

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago